First slide

Binomial theorem for positive integral Index

Question

The value of $^{n} C_{1} +^{n + 1} C_{2} +^{n + 2} C_{3} + \dots +^{n + m - 1} C_{m}$ is equal to

Moderate

A

$^{m + n} C_{n} - 1$
B

$^{m + n} C_{n + 1}$
C

$^{m} C_{1} +^{m + 1} C_{2} +^{m + 2} C_{3} + \dots +^{m + n - 1} C_{n}$
D

$^{m + n} C_{m} - 1$

Solution

$^{n} C_{1} +^{n + 1} C_{2} +^{n + 2} C_{3} + \dots +^{n + m - 1} C_{m} =^{n} C_{n - 1} +^{n + 1} C_{n - 1} +^{n + 2} C_{n - 1} + \dots +^{n + m - 1} C_{n - 1} = Coefficient of x^{n - 1} in (1 + x)^{n} + (1 + x)^{n + 1} + (1 + x)^{n + 2} + \dots + (1 + x)^{n + m - 1} = Coefficient of x^{n - 1} in (1 + x)^{n} [\frac{(1 + x)^{m - 1} - 1}{(1 + x) - 1}] = Coefficient of x^{n - 1} in \frac{(1 + x)^{m + n} - (1 + x)^{n}}{x} = Coefficient of x^{n} in [(1 + x)^{m + n} - (1 + x)^{n}] =^{m + n} C_{n} - 1$
Similarly, we can prove
$^{m} C_{1} +^{m + 1} C_{2} +^{m + 2} C_{3} + \dots +^{m + n - 1} C_{n} =^{m + n} C_{m} - 1$

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